# How do i convert this into a rectangular equation

• Jun 24th 2008, 08:06 PM
cityismine
How do i convert this into a rectangular equation
y = cos(2t), y = sin(2t);

t is greater than or equal to -pie and less than or equal to pie.
• Jun 24th 2008, 09:21 PM
mr fantastic
Quote:

Originally Posted by cityismine
y = cos(2t), y = sin(2t);

t is greater than or equal to -pie and less than or equal to pie.

$y^2 = \cos^2 (2t)$.
$x^2 = \sin^2 (2t)$.

Therefore $x^2 + y^2 = 1$.

Domain is [-1, 1] and range is [-1, 1] so it's the entire circle.
• Jun 25th 2008, 08:34 AM
Mathstud28
Just so you know, given

$x=a\cos(n\theta)$

and $y=a\sin(n\theta)$

You will always get a circle of radius a

$x^2+y^2=a^2\cos^2(n\theta)+a^2\sin^2(n\theta)=a^2$

and if you have

$x=a\cos(n\theta)$

and

$y=b\sin(n\theta)$

You will get an ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{a^2\cos^2(n\ theta)}{a^2}+\frac{b^2\sin^2(n\theta)}{b^2}=1$
• Jun 25th 2008, 09:10 AM
galactus
Quote:

-pie and less than or equal to pie.
Come on...this is math, not a bakery.(Wink)
• Jun 25th 2008, 06:30 PM
cityismine
The answer in the back of the book is: y=(1/2)x^2-1

Is the answer in the book wrong?
• Jun 25th 2008, 06:37 PM
Mathstud28
Quote:

Originally Posted by cityismine
The answer in the back of the book is: y=(1/2)x^2-1

Is the answer in the book wrong?

Yes

For we have that our equation is

$x^2+y^2=1\Rightarrow{y=\pm\sqrt{{\color{red}1-x^2}}}$
• Jun 25th 2008, 07:09 PM
mr fantastic
Quote:

Originally Posted by cityismine
The answer in the back of the book is: y=(1/2)x^2-1

Is the answer in the book wrong?

If the parametric equations were actually $x = \cos t$ and $y = \cos (2t)$ then the cartesian equation would be $y = 2x^2 - 1$ ........

If the parametric equations were actually $x = 2 \sin t$ and $y = \cos (2t)$ then the cartesian equation would be $y = 1 - \frac{x^2}{2}$ ........

Are the equations you posted correct?
• Jun 25th 2008, 08:29 PM
cityismine
Yes, I posted them correctly, I just double checked. I guess it's a typo in the book.